Final answer:
The general solution to the given differential equation 12y'' + 4y' - 5y = 0 is y(t) = C1e^(t/3) + C2e^(-5t/4), where C1 and C2 are arbitrary constants.
Step-by-step explanation:
To find a general solution to the given differential equation, let's first solve for the characteristic equation. The characteristic equation is obtained by replacing the derivatives with their respective differential operators. So, for our equation 12y'' + 4y' - 5y = 0, the characteristic equation is 12r^2 + 4r - 5 = 0. Solving this quadratic equation, we get two distinct roots: r = 1/3 and r = -5/4.
Therefore, the general solution to the differential equation is y(t) = C1e^(rt) + C2e^(st), where C1 and C2 are arbitrary constants. Substituting the values of r and s, we have y(t) = C1e^(t/3) + C2e^(-5t/4).